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# Heating Effect of Electric Current, Joule’s law: Solved Example Problems

Heating Effect of Electric Current: Solved Example Problems, Joule’s law: Solved Example Problems

## Heating Effect of Electric Current, Joule’s law: Solved Example Problems

EXAMPLE 2.27

Find the heat energy produced in a resistance of 10 Ω when 5 A current flows through it for 5 minutes.

Solution

R = 10 Ω, I = 5 A, t = 5 minutes = 5 × 60 s

H = I2 R t

= 52 × 10 × 5 × 60

=25 × 10 × 300

=25 × 3000

=75000 J (or) 75 kJ

EXAMPLE 2.28

An electric heater of resistance 10 Ω connected to 220 V power supply is immersed in the water of 1 kg. How long the electrical heater has to be switched on to increase its temperature from 30°C to 60°C. (The specific heat of water is s = 4200 J kg-1)

Solution

According to Joule’s heating law H = I2 Rt

The current passed through the electrical heater = 220V/10Ω = 22 A

The heat produced in one second by the electrical heater H = I2 R

The heat produced in one second H = (22)2 x 10 = 4840 J = 4.84 k J. In fact the power rating of this electrical heater is 4.84 k W.

The amount of energy to increase the temperature of 1kg water from 30°C to 60°C is

Q = ms ∆T (Refer XI physics vol 2, unit 8)

Here m = 1 kg,

s = 4200 J kg-1,

∆T = 30,

so Q = 1 × 4200 x 30 = 126 kJ

The time required to produce this heat energy t = Q/ I2R = 126 ×103 /  4840  ≈ 26 .03 s

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12th Physics : Current Electricity : Heating Effect of Electric Current, Joule’s law: Solved Example Problems |

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